Generator

The cheapest way out of being round

Rendered here at the parameters it defaults to, with every essay that calls it — which is the same list as the blast radius of changing it.

The cheapest way out of being round. A ring under uniform external pressure, drawn in its first four buckling modes with the pressure each one needs underneath it, in N/mm². The pressure has no direction: it stays normal to the wall wherever the wall goes, so it does work on any change of shape that reduces the enclosed area, and the ring buckles into whichever shape is cheapest. Bare, that is the oval — n = 2 at 3EI/R³ — and the modes rise as n² − 1, so three lobes cost 2.67 times as much. Nothing in the drawing prefers any orientation, which is the point — a column has an axis to buckle about and a ring has none.

7 essays call ring-buckling. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

Stability

The pressure that needs no direction

Every buckling problem in this collection has had a load with a direction — a column pushed along its axis, a plate along its edge, an arch by what is on it. A buried pipe has none. The pressure is the same everywhere, it stays normal to the wall as the wall moves, and it does work on any change of shape that reduces the area inside.

The free body that makes a hoop force a pressure times a radius. Half a ring cut along a diameter, with the pressure drawn normal to the wall wherever the wall is. Vertical equilibrium of the half ring is the whole derivation: the pressure acts over the projected width 2R whatever the shape of the arc, the two cut faces carry N each, so N = pR — 300 kN per metre here at 1 MPa on a 0.3 m radius. The result contains no wall thickness, no second moment, and no length along the pipe, which is why a hoop force is the one internal force in this collection that arrives with no lever arm attached to it. The stress does contain the thickness — 25 MPa at 12 mm — but the force does not, and a thicker wall carries exactly the same force at a lower stress.

Internal forces

The force that is only a radius

Every internal force in this collection arrives with a lever arm attached. A hoop force does not. Cut a cylinder along a diameter and the free body settles it in one line — pressure times radius, with no thickness, no second moment and no length in it — which is why a tank wall is thin and why its worst hoop force is not at the bottom.

$The lining that carries less for being weaker. Bending moment and hoop thrust in a circular lining, against the lining's own bending stiffness, both as fractions of the free-ring values. The ground arrives already stressed — 500 kPa vertically and 300 horizontally at K₀ = 0.6 — and the difference between them tries to squash the hole into an ellipse. A lining stiff enough to refuse absolutely collects the whole distortion pressure, p₂R²/3 = 300 kNm/m; one flexible enough to go with the ground collects nothing, because there is no curvature change left to resist. The thrust is the flat line: it comes from the mean stress rather than the difference, so it does not move at all. Putting 8 joints in this ring drops the moment to 34% of the solid one and leaves the thrust exactly where it was, which is why a segmental lining is jointed and why the intuition carried over from a beam is inverted here.$ Structural form

The lining that is stronger for being weaker

A tunnel lining is not loaded. The ground arrives already stressed and the hole wants to squash into an ellipse; the lining's only job is to refuse, and how much moment it collects depends entirely on how hard it refuses. Make it stiffer and it takes more. Make it flexible — put joints in it, make it thin — and it takes almost none, while the hoop thrust it carries does not move at all.

Two identical pipes, and one carries three times the other. Load per metre on a buried conduit against the depth of cover, in trench widths, with the weight of the prism of soil directly above it drawn between them. A conduit laid in a narrow trench is stiffer than nothing and softer than the sides: the backfill settles relative to the undisturbed ground, the friction on the trench walls acts upward, and the conduit gets 64% of the prism. Lay the same conduit on the ground and build an embankment over it and it is now stiffer than the fill beside it, the interior prism settles less, the friction acts downward, and it gets 172% — a factor of 2.71 between two pipes with nothing different but which way the ground moved. The equation is Janssen's, the same one a silo wall obeys, with a trench for a silo; both curves start on the prism line, because with no depth there is no shear to redistribute anything. This is why a flexible pipe is buried rather than a rigid one: making the conduit weaker moves it down the page.

Equilibrium

The pipe decides what the soil weighs

A buried conduit is not loaded by the soil above it. It is loaded by whatever share of that soil the relative movement leaves it — and which way the shear on the sides of the prism acts depends on whether the conduit settles more or less than the ground beside it. Two identical pipes under identical fill, one carrying two thirds of the prism and one carrying nearly twice it.

The tube flattens because of the bending, and then cannot carry it. Moment against curvature for a long tube of radius 300 mm and wall 4 mm. Compression on one face and tension on the other are both directed along a curved line, so each produces an inward transverse pressure and the circle is squashed into an oval by the bending it is carrying. That reduces the second moment, so the curve bends over and reaches a limit point — no bifurcation, no imperfection, nothing to be sensitive to. It arrives at an ovalisation of exactly 2/9 for every tube of every size in every material, at 1018 kNm, where the secant stiffness has fallen to 67 per cent of the undeformed value and the tangent stiffness is zero. The relaxed path reproduces the closed form to 0.004 per cent.

Stability

The tube that flattens itself

Bend a tube and the compression on one face and the tension on the other are both running along a curve, so both push inward. The circle becomes an oval, the second moment falls, and the moment–curvature curve turns over at a limit point that needs no imperfection, no bifurcation and nothing to be sensitive to.

The limit moment goes as the square root of the ring's stiffness, and pressure is part of it. Brazier's limit moment for a steel tube of radius 600 mm and wall 2.5 mm (r/t = 240), as a multiple of its value with no pressure, against the pressure as a multiple of the ring's own buckling pressure, pc = 3D/r³ = 4.07 kPa; positive inside. The dots are the peaks of the moment–curvature path found by minimising the energy at each curvature; the line is √(1 + p/pc). Half pc outside leaves 0.71 of the bending capacity, and the capacity is gone at pc; pc inside gives 1.41, three times pc 2.00. The pressure enters only by adding to the ring's stiffness against the oval, (3D/r³ + p) where it had 3D/r³.

Stability

Three kilopascals that halve a tube in bending

A long thin tube in bending flattens until its section can carry no more, at an oval of exactly two ninths of its radius. Put a pressure on it and nothing about that shape changes. What changes is how much bending it takes to get there — as the square root of one plus the pressure over the ring's own buckling pressure — and for a steel tube 1.2 m across with a 2.5 mm wall that unit is 4 kPa. Three kilopascals of suction halve its bending capacity; a fifth of an atmosphere inside doubles it and hands the tube to another failure entirely.

Which way the oval lies decides whether the pressure helps or harms. The limit moment of a steel tube of radius 600 mm and wall 2.5 mm (r/t = 240), whose ring buckles under 4.07 kPa, over its limit with no pressure (796 kN·m), against the angle between the bending plane and the diameter a non-uniform pressure pushes in on, for second harmonics of 0.10, 0.25 and 0.50 of that pressure; lines the closed form, dots the minimised energy. At 0.10, 0.93 aligned and 1.08 across; at 0.25, 0.82 aligned and 1.19 across; at 0.50, 0.65 aligned and 1.40 across; at 45° the oval does not touch the limit at all.

Stability

The oval the pressure brings

A uniform pressure changes how stiff a tube's ring is against the oval that bending squashes it into, and nothing else: the tube still gives way at a flattening of two ninths of its radius. A pressure that is larger on one diameter than the other has an oval of its own, and that oval does work on the bending's directly. Lying in the bending plane it has done some of the flattening before the bending starts, and a pressure difference a quarter of the ring's own buckling pressure takes a fifth off the tube's bending capacity; lying across it, it adds a fifth. At forty-five degrees it does nothing at all. And the two ninths turn out to have been a property of a round tube.

The library, page 5 of 7 — where ring-buckling sits